Find empirical formula from percentage composition: C=40% H=6.7% O=53.3%

Question

Find the empirical formula of a compound with the following percentage composition: C = 40%, H = 6.7%, O = 53.3%.

Solution — Step by Step

This converts percentages directly into grams:

Mass of C = 40 g
Mass of H = 6.7 g
Mass of O = 53.3 g

Divide each mass by the element’s atomic mass:

n(\text{C}) = \frac{40}{12} = 3.33 \text{ mol}

n(\text{H}) = \frac{6.7}{1} = 6.7 \text{ mol}

n(\text{O}) = \frac{53.3}{16} = 3.33 \text{ mol}

Divide all values by the smallest value (3.33):

\text{C} : \frac{3.33}{3.33} = 1

\text{H} : \frac{6.7}{3.33} = 2.01 \approx 2

\text{O} : \frac{3.33}{3.33} = 1

The ratio C : H : O = 1 : 2 : 1.

\boxed{\text{CH}_2\text{O}}

The empirical formula is CH₂O. The molar mass of CH₂O = 12 + 2 + 16 = 30 g/mol.

Note: This is the simplest formula. The actual molecular formula could be CH₂O (formaldehyde), C₂H₄O₂ (acetic acid), C₃H₆O₃, C₆H₁₂O₆ (glucose), etc. — all have the empirical formula CH₂O.

Why This Works

The empirical formula gives the simplest whole-number ratio of atoms in a compound. We convert percentages to moles because moles directly represent relative numbers of atoms. Dividing by the smallest mole value normalises the ratios to their simplest form.

The assumption of 100 g is a clever trick: it makes the calculation cleaner without changing any ratios (percentages are ratios, so scaling to 100 g preserves them).

This four-step method works for any percentage composition problem:

Assume 100 g
% → mass (g) directly
Mass ÷ atomic mass → moles
Moles ÷ smallest → ratio (round to nearest whole number)

If you get a ratio like 1 : 1.5, multiply all by 2. If you get 1 : 1.33, multiply by 3. Don’t round until the step where you check for whole-number ratios.

Alternative Method

You can also express the calculation in tabular form for board exam presentation:

Element	%	Atomic mass	Moles ( $\%$ /at. mass)	Ratio
C	40	12	3.33	1
H	6.7	1	6.70	2
O	53.3	16	3.33	1

Empirical formula = CH₂O.

Common Mistake

Students sometimes round moles prematurely (e.g., writing $n(\text{H}) = 6.7 \approx 7$ before comparing) and get the wrong ratio. Only round to the nearest whole number in the final ratio step, after dividing by the smallest value. Also, students sometimes forget to check whether the percentages add up to 100% — if they don’t, there may be a fourth element present (often nitrogen or a halogen).

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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